HP vs Torque vs Fuel Efficiency

[sendler]

Originally Posted by miprofessor

http://www.lainefamily.com/hp.htm

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Good link. This could become another good thought experiment such as some other topics here did.
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http://www.cbr250.net/forum/cbr250-p...er-half-3.html
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An engine torque curve and horsepower curve are really just two different ways of looking at the same thing. Power is just torque times time. People like to look at the torque curve to get an idea of what is going on below 5,000 rpm because it has better resolution on the graph due to the horsepower curve not yet having much multiplication from the low rpm. The power doesn't yet look like much and the torque looks much bigger but they are none the less showing the same information. And power is the real metric of how much work can be done. Torque can be changed any way you want with gearing but that never changes the power at the wheel because you are trading increased rear wheel torque for decreased wheel rpm. Increasing the rearwheel torque WILL make the bike accelerate quicker but when you get to the steady state top speed for any given load such as the wind on a flat top speed run or gravity on a big hill, the gear that will be fastest is the gear that puts the rpm at the power peak. Not the torque peak. And not the redline. Even a drag strip run which is all about acceleration and using the most torque multiplication, will be the quickest on a vehicle like the CBR250R when shifting the higher gears at an rpm band that is much lower than the redline and is centered on the power peak. As was generously explained in the other thread by DieselMaxPower
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The OP's link is a little too simplistic when it comes to fuel efficiency. It is possible for an engine to take in the same amount of fuel and air (disregarding the much higher losses at high rpm) at 8,000 rpm as it is at 4,000 because of the throttle plate's restriction and resulting intake vacuum. The real reason for generally better fuel economy at lower rpm is from less frictional, reciprocating, and thermal losses than the higher rpm. And from being closer to the first torque peak.
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The first torque peak usually shows where the engine is operating most efficiently as set up by the cam timing, intake runner tuning, ect versus linear piston speed losses, flame front travel and combustion gas pressure curves, ect,ect. A Brake Specific Fuel Consumption chart (Which plots 4 parameters; engine rpm along the bottom, the torque that is produced along the right, increasing engine load which can be roughly increasing throttle position up the left, and finally the topo values for how much fuel is being used.) is the only way to really know the engine's best efficiency range. They usually show the best efficiency at just off of full throttle on either side of the first torque peak. Which is the bad news for fuel economy. Most engines are way too powerful to operate at full throttle for more than a few seconds without going way too fast. So small engines with barely enough power, like our 250, are more often operated near their most efficient range than an engine with plenty of extra power. The only way to get the bigger engine into it's efficient range is to pulse with a high throttle opening and then glide with the clutch in, and then pulse again. But this an advanced hypermiling technique that is not for everyone. Better to stick to a CBR250R, PCX150 scooter, or new Ford Fiesta.
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Thermal losses, pumping losses, friction losses and powerband tuning, among many other factors, all come into the design equation. Using the smallest combustion chamber and the fewest of them minimizes the area which can lose heat (wasted fuel) to the head. But too small of a bore reduces the force of any given combustion pressure because it offers less area to work on. So the tendency for engine design to have a square (equal bore/ stroke) design. Slightly longer stroke than bore will be more efficient as it will move the torque peak to a lower rpm because of the longer crank arm. Interestingly, a square single will have less combustion chamber area to pick up heat than a square twin of the same displacement. Multiply it out for yourself. It's cool. And so is the CBR250R's engine. I'm sure you have noticed that the engine and radiator don't throw off much heat compared to your other bikes. Air cooled engines can run cylinder head temps of 350F/ 175C which, compared to liquid cooled engines which are limited to 100C to prevent boiling, waste even less heat.
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Heat out the exhaust is another waste. Which can be reduced with Atkinson cycle engines
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http://en.wikipedia.org/wiki/Atkinson_cycle
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(a fancy way of saying that the cam left the intake valve open to long.) which seek to have no pressure remaining in the cylinder by the time the piston reaches the bottom. And they also reduce pumping losses with lower intake vacuum. At the expense of power versus displacement which never looks good to bench racers reading spec sheets.
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Honda motorcycles is fighting for better fuel fuel efficiency in the big bike west with the ultra long stroke CBR500 platform and ultra low rpm tuned NC700. Either one of which can almost match our already exemplary CBR250R. BMW has always been a fuel efficiency leader. The Ninja300 is another sporty and all around good bike taking a step in the right direction with longer stroke and longer gearing. Some small 110cc air cooled bikes which are popular in India can sell for $1000 and get 150 mpgUS at 40 mph.

[sendler]

Here is a sample BSFC map from another vehicle and a chart of the optimum shift rpm's of the CBR250R. The shift points were charted by DieselMaxPower from a dyno printout to show when the engine torque as multiplied by the drive train at the wheel would be greater in the next gear. Notice that the top three shifts are giving the most average rear wheel torque and power when shifting at about 9,300 rpm. Redline is 10,500.
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[choneofakind]

Interesting read. Thanks for sharing this with us.

Originally Posted by sendler

The shift points were charted by DieselMaxPower from a dyno printout to show when the engine torque as multiplied by the drive train at the wheel would be greater in the next gear. Notice that the top three shifts are giving the most average rear wheel torque and power when shifting at about 9,300 rpm. Redline is 10,500.

This is why I got faster 0-60 runs when I shifted before 12,500 instead of waiting all the way until 14,000+. I've just never been able to put it into words. Looks like I should try shifting at 11,000 or so from 1st to second. I know they're different engines, but there's still going to be a lower shift rpm for the lower gears.

[sendler]

The spread between the 1-2 is usually much bigger than the higher gears so 1st is the one gear on most bikes that will benfit by running up to redline so as not to drop down so far out of the power band in 2nd. You would have to look at a dyno for your bike and center all the gears speed rpms on either side of the power band. Gearing commander can show you the speed vs rpm in each gear for any bike you load up.

[choneofakind]

True. Still a neat read though.

[sendler]

No gear heads here?

[sendler]

Here is a reply from another forum which shows the classic and incorrect view of torque vs horsepower.

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"Related, but not the same. I have a Dodge Ram 3500 Cummins diesel. A BMW sportscar will smoke it in a race. It has more HP. But lets see it pull 25,000 pounds of round hay bales at 70 MPH up a steep grade! My diesel has a redline of just 3,375 RPM or so. It makes all it's power from only 1,600 RPM and up. It only has 230 or so HP, but it has gobs or torque, something like 460 ft/lbs@1600."

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Thanks for the reply. I wondered if there were any gear heads here that would want to discuss this. Yours is the common view based on intuition but it isn't quite right. This would actually make an amusing episode of Top Gear. Car vs truck: TOWING!

I found an interesting dyno test where the Ford 6.7 Powerstroke engine made 700 foot pounds and 350 hp.

The truck will be better to tow with since it has greater torque and power down low. So when you take off from a light, it is already in the power band at the torque converter stall rpm. It is easier to access it's POWER band right off the line. The car is tuned for high power closer to redline. At the expense of not having as much torque or power at lower rpm. So even in first gear it's rpms are to low to have the torque or POWER to yank the trailer to get started.

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But, if you were to race two trucks and two trailers at the drag strip, the driver who makes his shifts on either side of the engines power peak would beat the other driver who shifted on either side of the torque peak. Power is doing the work of moving the trailer. And the rear wheel torque of the winning truck will avreage out to be HIGHER than the earlier shifting truck who used his engine at the range of greater torque. The slower truck sent more torque to the trans but the faster truck stayed in a lower gear more often and used a greater POWER band to get greater rear wheel torque, and hence power, at any given speed.

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Here is the crazy part.

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Once it finally got moving so it could use it's gears at the power peak, a 400 hp car will pull the trailer up the hill at a faster speed than a 350 hp truck. Even though the car doesn't make as much torque as the truck anywhere. Power is doing the work. A lesser engine torque can be multiplied with gears but the power at the wheels is always the same as the engine is putting out.. The trans is trading higher wheel torque for lower wheel rpm. The power at the wheel is the same as the engine at any gear. And higher power will do more work than lower power. Most towing vehicles have giant engines which are tuned for low rpm so they have tons of torque, and so, power, down low to yank with, and plenty of peak power to keep moving up the hill from the large displacement. But it is always the amount of power that is doing the work.

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This is interesting. The first google search I made came up with test data on a dyno which states that the rear wheel power was the same in any gear. If anything, the power showed to be slightly higher in 5th than in 3rd. Must be the dyno is not really that linear at different speed ranges because that is the complete opposite from what most people would think where they would guess that the lower gear would give the most rear wheel power. And in reality the power would be the same discounting different losses in the trans at different gears.

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http://www.dieselpowermag.com/featur...diesels_built/

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"We strapped down the 2011 ½ Ram 3500 dualie to the rollers first. The guys at ATS made three dyno pulls with the new engine, one in Third, one in Forth, and one in Fifth gear. The results were surprising.

Regardless of which transmission gear we tested the trucks in, the power rankings were all the same"

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Here is an interesting quote by DieselMaxPower at the Honda site which explains the conundrum of the power at the rear wheel not being multiplied by using a lower gear even though the vehicle does accelerate much faster. An example would be The CBR250R with an engine power peak at 8,700. If you are rolling along at a steady speed in first at 8,700 rpm, and then crack full throttle, the bike will accelerate violently. We can feel this very obviously. Now do the same in top gear. The bike is felt to start going faster but the rate of acceleration is much, much less. Even though we used the same amount of power from the engine and at the wheel. The acceleration is much greater with the lower gearing but we have put the same amount of power to the ground over time as in the higher gear. There must be some quantity that is increasing at the same rate in either gear (disregarding the higher resistance of speed to wind drag in top gear).

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Ready?

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The ENERGY of the system is increasing at the same rate in either gear. But we are very bad at perceiving energy so it seems like top gear does very little and first gear is doing much more with the same amount of power.

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Example of how bad we are at perceiving differences in energy:

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Cruising along at a steady speed in a car so you can't feel the wind, 10 mph feels pretty much the same as 70 mph. But 70 mph has much more energy and would be a very bad time to run into a wall.

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From DiesleMaxPower

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http://www.cbr250.net/forum/102283-post75.html

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"Torque doesn't do one thing while horsepower does another. Torque is a measure of what's happening to the bike. So is horsepower. It depends on what you want to look at. Torque can give you a way to look at the accelerations. This is great for most humans because we can perceive things like position, speed, and acceleration. Horsepower lets you look at energy. Humans don't have a good way to perceive energy, except for maybe temperature changes which we don't include. When we analyze the dynamics of the bike we look at the speed of the bike (thats the momentum side) and the kinetic energy (thats the energy side). In the simple analysis we're doing now the two are so simplistic that they will always be consistent with one another. It doesn't matter if you decide that torque moves the bike or horsepower. The reason I like to look at HP because the math is much easier, as we've shown with our torque charts. You don't need to transform it through gears."

[greenaero]

Sendler, Great post and information! Thanks for sharing this. The BFSC map clearly shaws that the the best fuel economy corresponds perfectly with the the first torque peak. This is the rpm to shoot for maximum fuel economy.

[subxero]

Originally Posted by sendler

The spread between the 1-2 is usually much bigger than the higher gears so 1st is the one gear on most bikes that will benfit by running up to redline so as not to drop down so far out of the power band in 2nd. You would have to look at a dyno for your bike and center all the gears speed rpms on either side of the power band. Gearing commander can show you the speed vs rpm in each gear for any bike you load up.

Just going by how both my cars and my bikes pull. I feel like the higher revs of 1st gear do very little for increasing actual speed. So my acceleration in these higher revs is small compared to the lower revs of first gear while other gears continue to pull up through the higher revs, incase of the ninja250 ~ 11,500 rpm.

I know in my Civic Si before i sold it, i would never run 1st gear out high, It would take off good but actual acceleration seemed to die early so i would always shift out early to hit second then rev match with my clutch to get second gear to pull until i hit the magic rpm of 6500 where vtec kicks in, then after that shift at redline so your next gear would fall in the beginning of the powerband 6500rpm

I feel my ninja pulls similar, could be a jetting problem i suppose but first gear feels flat up top IMO, would rather get out early and rev match second gear to get rpms up. IDK, Could peak torque be at lower RPM in first gear compared to other gears? I guess i could be imagining things, since all gears tend to run a little flat at very high end of RPM range, is it just more noticeable in 1st? Would be interesting to see dyno runs of all gears, overlaid, most people don't do 1st gear pulls though.

[sendler]

A dyno power curve will have the same values in any gear. They tried it on the trucks in the link above and three different gears gave essentially the same power curve.
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Shifting either side of the power peak, which for most vehicles will be redline for 1st to keep the wide spread from dropping down so far out of the power band. The higher gears may be better to shift earlier on an engine like the CBR250R since they are closer and the engine is tuned for fuel economy with a low rpm power band.

[subxero]

Originally Posted by sendler

A dyno power curve will have the same values in any gear. They tried it on the trucks in the link above and three different gears gave essentially the same power curve.
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Shifting either side of the power peak, which for most vehicles will be redline for 1st to keep the wide spread from dropping down so far out of the power band. The higher gears may be better to shift earlier on an engine like the CBR250R since they are closer and the engine is tuned for fuel economy with a low rpm power band.

Yes i read that and understand what they said but none of which answered my questions.

Here is another question related to all that before i get to my other point.
So in same car Civic Si, I could easily redline 5th at around 125mph 8500rpm the whole time the car pulled great but, shift to 6th rpms start at 6k or so and the car would not pull at all, you would pretty much have to be going down hill to get above 130mph in 6th not even close to redline but still in good RPM range for power. If both gears make the same power, why can i pull strong to redline in 5th, but 6th won't pull at all in the meat of the power band? When 5th gear redlines out it still is pulling hard, even up hill.


Back to my original conundrum
Like i said from my experience 1st gear seems to run flat on most vehicles up top noticeably, but why? Again does it just have to do with the gear and it is more noticeable because the gear is run in such a short speed range? just for numbers sake say 0-25mph (i do not know exact numbers for ninja250) before your are completely redline, where as other gears can be run in a broader speed range and out of their optimal range giving you a better feel for difference of power throughout the revs. But in 1st gear, it is run from 0-25mph, no more, no less and you don't get a feel for that lack of power in the low revs because you are out of it quickly, into power band, then it flattens out at the top just as quickly giving you that flat top end feel?

This is all pure speculation i am simply looking for a reason to something i feel when i ride or drive.

[sendler]

Originally Posted by subxero

I could easily redline 5th at around 125mph 8500rpm the whole time the car pulled great but, shift to 6th rpms start at 6k or so and the car would not pull at all

Your car makes more power at 8,500 than it does at 6,000.

[sendler]

here is a reply from another site and my answer.
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"Here's the problem... you can't disregard air resistance at higher speeds. That's like saying granite would float in water if you disregard gravity.

In top gear, you use a significant amount of your power due to wind resistance and friction. Only the remainder is used to accelerate the vehicle. Since you have less available power to accelerate the vehicle, you will accelerate slower.

It is also more "violent" in low gears because you have more torque at the wheels which is more likely to get the tire to lose traction or for the reaction to lift the nose of your vehicle upwards."

But consider this: What about the turbo diesel trucks in the link which all have a basically FLAT power curve. Could you then choose from 3rd, 4th, or 5th which would give the same exact power and road speed? Just a different rpm and torque multiplication. And expect the exact same level of acceleration? Crazy topic huh. I think 3rd will accelerate much harder.
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[subxero]

Originally Posted by sendler

Your car makes more power at 8,500 than it does at 6,000.

The peak power was probably at 7k rpm i can't remember exactly but power output from 6k to 8500 should not have been all that different that it could not pull much at all out of 6th

[sendler]

My last example with the trucks was a bad one. And my statement that 3rd would pull harder than 5th given the same speed and power with just a different rpm along the super flat power curve was wrong. The acceleration would be the same. The starting speed was the same in each example and so was the power being applied. And so was the rear wheel torque in either gear. More on this coming right up.
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Even though I was wrong in that statement, it helped me get back to the premise I am trying to prove.
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First gear accelerates at a much higher value than 6th. And not because of air resistance at speed in 6th.
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Because we are comparing changes in energy.
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Lets use horsepower hours for energy to keep the terms we have already been using. Or, horse power seconds to put into the appropriate scale. If I have to come up with exact equations for mass and speeds to support my point, I will. But I think we can get the idea from the concepts, without resorting to using exact numbers which will waste a lot of time to throw together.
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We have a CBR250R and the area under the power band we are using as we shift from the high side of the power peak to the low side with each gear is the same from 9,500 rpm to 7,500 and averages 20 hp. Let's say we start with 20 horse power seconds to begin with and 226kg at 6.7 meters/ second. So we pin the throttle while cruising at 7,500 rpm, 1 second later we have added 20hp seconds of energy to what ever we had before. Now we are up to 9.47 m/s and are carrying 40 hps. At the end of 2 seconds we will be carrying 60 hps and 11.6 m/s. We have doubled our energy vs the starting point in the first second and added another 33% to that in the second second. These are big changes of energy vs time. And big changes of speed vs time. So, a high acceleration. .276 G over the first second and .217 G over the second second. Now we continue to accelerate and shift up through the gears each time we reach 9.500 rpm until we hit 5th at the ten second mark. We now have 220 hp seconds of energy and 22.22 m/s. One more second later we have 240 hps and 23.21 m/s. We only gained 1 meter per second which is .1 G of acceleration. Less than half the acceleration we had between the first and second seconds of the run when the speed was low.
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This has nothing to do with wind yet which will indeed have a big effect on such a small amount of horsepower and acceleration. It is because the energy we are putting in is just increasing linearly with time but the speed is squared on it's side of the equation.
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And this brings me to the real revelation I had from the truck's dynos and 3rd accelerating the same as 5th. Power is the metric of how fast you can gain speed as I have been saying. And in the end, the torque multiplication of the different gears doesn't really do anything to make something pull harder other than to match the engine power band to the needed wheel speed.

[subxero]

^ I think i came to the same conclusion about why 1st gear feels so flat up top but in a much less mathematical and complicated way

In short. 1st gear feels flat up top well, because it is most gears in most vehicles are flat in very high revs.
Much like Sendler was getting at. You feel acceleration, change in velocity over a set time frame. It doesn't take a genius to realize you accelerate more from 0-25mph, more than you do from 60-85mph. delta Velocity is still the same 25mph but the time is significantly different when achieving this change in V so the acceleration you feel is also significantly different. Because of this first gear will tend to feel flat early and up top and not seem like it is pulling or doing you any good but in fact it still is.

Yes the gear runs flat up top...all do, you just notice it more in first due to higher rate of acceleration and then the change in this acceleration when the power flattens out.

Anyone please feel free to chime in and say "DUH!!"

ehhh my head hurts....

[Samer]

Originally Posted by sendler

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An engine torque curve and horsepower curve are really just two different ways of looking at the same thing. Power is just torque times time.

First, thanks for posting this as I have seen a lot of confusion about it over the years.

I would just like to nit pick a bit and correct your second sentence: Power is torque times rotational speed.

Also, one more thing, in case there may be confusion about this. In the equation:

Power (HP) = Torque (lbf*ft) * RPM (rev/min) / 5252

The number 5252 is nothing magical. It is not a universal truth that should be held with great admiration like say for example, the number pi, which is a universal truth. The number 5252 is completely arbitrary and is just a unit conversion thing related to the Enlish system of units.

Just in case anyone is interested, I actually went through the derivation of it now in order to demonstrate this:

Power in general is torque times rotational speed. The units would be:

Power = Torque * Radians / Time

In English Units:

Power (HP) = Torque (lbf * ft) * RPM (rev / min) * (2*pi rad / 1 rev) * (1 min / 60 sec) * (1 HP / 550 ft * lbf / s)

Because the convention is to divide by a number to calculate power, you would take the reciprocal of these units:

550 * 60 * (1/2*pi) = 5252.11... which is approximately 5252

Anyway, I just wanted to show it is just a dumb unit thing. If we were to use SI (Some call it Metric), things would be a lot less messy. We would still have to do the revolutions per minute to radians per second thing just because everyone is calibrated to RPM, but that is it.

A Watt of power is just a Newton * Meter / Second

A radian has no unit.

[sendler]

Still at it and working some of the bugs out of some misstatements earlier. The highest rear wheel torque will give the highest acceleration.
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The faster you go, the less torque you have as you shift up through the gears if power is constant. This is intuitive but it is also predicted by
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P=F*V
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This holds true according to theory and measurement even for a variable pitch airplane prop which is like a cvt transmission. Check out the chart that shows the thrust of a certain variable pitch prop at constant horsepower and vs increasing speed which is calculated from
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Thrust = ( HP * eff * 375 ) / air speed (mph)
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and looks like this:.
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The lines rise at first because a real world prop has poor efficiency at low speeds. Efficiency actually rises throughout the increasing speeds from 65% at 100 mph to 85% at 240. Otherwise the slope of the thrust drop off would be even steeper.
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The physics we need to understand are really very simple. Reciprocating engines make Torque and when looked at by multiplying that by the Revolutions (How many "Torques") in a certain amount of Time, you get the Power of the engine.
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hp=Torque * RPM/ 5252
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Rear wheel Torque is Force on the road. But it is interesting to note that the highest rear wheel Torque will occur, ie with a constantly variable transmission, which is set to the Power peak of the engine. Not the Torque peak. Due to the fact that it will be using more torque multiplication (lower gear) at the higher rpm of the power peak.
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Force (from the rear wheel Torque) times the Distance the bike moved is Work.
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Power is the measure of Work vs Time. If you do the same amount of Work in a shorter Time, you have used more Power.
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P=W/T
Work=Force*Distance so
P=F*D/T and D/T is Velocity so
P=F*V
Given the same Power from the engine, The Force at the rear wheel must be less as the Speed between the road and the bike increases.
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F=M*A
Acceleration = Force/ Mass
F, and therefore A, must be less and less as the speed difference between the driven wheel and the road, or the airplane propeller and the air, is increasing.
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So this is the "Captain Obvious" over complicated way of saying the bike will pull harder in 1st than in 6th.
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Mass ejection engines such as a rocket may be different. I'm having a tough time with this and I must admit that I intuitively think that for vehicles propelled by ejected mass, rather than a driven member that is coupled to a stationary medium like the ground or air, or water, that the speed that is valid for the calculation of Force to the frame of reference of the passenger, is the speed difference between the gases and the nozzle. Not between the gasses and the Inertial Frame of Reference of the Universe. Which might simply indicate that the term constant Power doesn't apply to such engines which will have fairly constant force at the nozzle despite the Velocity of the rocket. I'm having a hard time finding anything to contradict this. I was hoping to find some model rocketry pages with measured thrust data to compare to the well documented bench testing of the various engines which could show if speed of the rocket comes into play as it does with the bike.
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ThrustCurve Hobby Rocket Motor Data
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Some interesting info on rocket engines I dug up: The efficiency of a rocket engine changes, starting at near 0% on lift off because the only thing moving is the ejected matter. And increases with increasing speed up to the best efficiency when the velocity of the exhaust gas as referenced to the nozzle is equal and opposite to the Velocity of the rocket referenced to the Inertial Frame of Reference. The rocket is flying away from the ejected mass at mach 10 or whatever and leaving the mass which is exiting the nozzle at mach 10, essentially motionless in space behind it. And a rocket engine can continue to supply force to a vehicle that is traveling faster than the exhaust speed at the nozzle. But with decreasing efficiency as now the exhaust will again have speed and momentum of it's own left over and in the same direction as the vehicle.
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Rocket engine - Wikipedia, the free encyclopedia
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Rocket - Wikipedia, the free encyclopedia

[sendler]

Here is the last piece we needed tying Power to Energy as posted by DieselMaxPower:
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"I'll add some info, even though you've basically got it. So a simplified equation for your last post is

P = M*A*V

It took several relationships, and if you actually did it through torque you would have several factors of 2*pi and the wheel radius to carry around and eventually cancel. Now I know most people don't know calculus, so you will have to trust me on this:

E = 1/2 M V^2
d(E)/dt = M V (d(V)/dt)

d(E)/dt is power, and d(V)/dt is acceleration. So check this out
P = M V A

You get the EXACT same equation using calculus. And acceleration is:

A = P/(MV),

so as you go faster, your acceleration decreases (assuming constant P). This doesn't really add too much to your current understanding, but if you want to more advanced stuff it will really help. For instance, we may want to take into account the force it takes to spin the wheels. This gets so ridiculously tedious looking at forces. Instead, add it to the energy equation

E = 1/2 M*V^2 + 1/2 I omega^2
d(E)/dt = M*V*(d(V)/dt) + I omega (d(omega)/dt)

I is the mass moment of inertia and omega is the angular velocity of the wheel, and now d(omega)/dt is the angular acceleration of the wheel. In the simplest case (tires aren't slipping on the road, or slip by a constant percentage), d(omega)/dt is proportional to A:

P = M*V*A + I*omega*c*A
P = A*(M*V+I*omega*c)
A = P/(M*V+I*omega*c)
A = P/(M*V+I*V*d)

where c and d are constants. We can change omega to V by the same argument that we could change d(omega)/dt to A. So acceleration decreases with increased velocity (assuming constant power), but now we get a more realistic estimate. Some of the power produced goes into spinning the wheels.

Rockets get into a whole new mess, which I would actually say makes more sense by looking at momentum instead of energy. If you're interested, the Reynolds transport theorem for momentum should give you some good information. Most simple problems don't need calculus, even though the theorem in its purest form needs it. As a bonus, looking at model rockets is one of the most common simple problems."

[sendler]

C*V*T = Obvious
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http://en.wikipedia.org/wiki/Continu...e_transmission
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A continuously variable transmission makes the concept of combining the equations for power and energy very obvious.
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Power = Mass * Acceleration * Velocity
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When you grab full throttle with a CVT, the transmission is controlled to change down to a lower ratio and wind the engine up to the power peak rpm. Where the rpm stays for the whole run until you let off the throttle. The CVT then continuously varies the gear ratio longer and longer to keep the same engine rpm while the vehicle Velocity is increasing from the Acceleration.
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The the throttle is held fully open by the rider and the engine will be kept at the same rpm by the transmission so the power from the engine is the same throughout the run even though the Velocity is increasing. The transmission is trading torque multiplication for speed as it moves up through it's range of gearing by squeezing the belt up in the front to make the engine pulley bigger and the wheel pulley smaller. The Power at the wheel is the same the whole time but there is less and less rear wheel Torque as Velocity increases.
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P = M*A*V
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Power stayed the same. And Mass stayed the same. So as the Velocity increases, the Acceleration and the Velocity must trade. One goes down if the other one goes up. Because the Energy that is added by the engine Power over time is linear. 1, 2, 3, 4. Going back to the Energy equation
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E = 1/2M*V^2
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Energy would have to increase exponentially to keep the same Acceleration as Velocity increases. 1, 4, 9, 16.

[sendler]

There is another misconception about gearing that is related to energy. Motorcycle guys often change the final gear ratio by changing sprockets since it so easy and cheap to do. I went 15% longer for better fuel economy. Some guys go shorter for better acceleration and to get what they believe will be closer gear ratios. The interesting thing is that the rpm change between gears stays exactly the same either way. If you run 1st to redline at 10,500 and the rpm drops to 6,600 when you grab 2nd, that difference will remain the same regardless of what changes you make to the final drive. You would have to change the gear sets in the transmission to make the rpm drop to be closer between gears. But why do the gears FEEL closer with a lower final drive and FEEL so far apart with my long gearing when you are getting the same rpm drop as before? Because each shift point is now closer together in Energy with the shorter gearing.

[sendler]

A racer brought up a good point about going shorter on the final drive. You can make the rpm drop closer together by going shorter in the final drive. Throw away 6th by making it more like what 5th was. Make 5th into 4th, 4th into 3rd, ect. Now you are more often using one gear up from what you were using before, where most transmissions group the ratios closer together at the top.

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I cruise long distance on the highway so I use 6th (can't throw it away). And 7th (what was 6th before I went longer on my sprockets).